Question: $f\,^{\prime}(x)=-3e^x$ and $f(1)=12-3e$. $f(0) = $
Solution: Finding $f(x)$ We have $f'(x)=-3e^x$ and we want to find $f(x)$ : $\begin{aligned}f(x) &= \int f'(x)\,dx \\\\ & = \int (-3e^x)\,dx \\\\ & = {-3e^x} {+ C} \end{aligned}$ Finding $ C$ Goal: We need to find $ C$ such that $f(1)=12-3e$. Here's what we get when we plug in $-1$ : $\begin{aligned}f(-1)&={-3e^{1}} {+ C}\\\\ &={-3e} {+ C} \end{aligned}$ We are given that this must equal $12-3e$ : $12-3e = {-3e} {+ C}$ Solving the equation gives us ${C=12}$. Finding $f(0)$ Now, we have that $f(x)={-3e^x} {+ 12}$. Let's find $f(0)$ by plugging in $0$ : $\begin{aligned}f(0)&=-3e^0 + 12\\\\ &=9 \end{aligned}$ The answer $f(0) = 9$